Solution

Let's define "magnification" as
$k = \frac{s'}{s}$ (1)
where $s'$ is the image distance from len and $s$ is the object distance from camera.
The rate of change for the object-to-len's distance is
$\dot{s} = -v$ (2)
it has negative sign due to the object goes towards the camera.
The relation with focal distance is given by
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ (3)
Differentiating (3) respect to time,
$\frac{\dot{s}}{s^2}+\frac{\dot{s'}}{s'^2} = \frac{\dot{f}}{f^2}$ (4)
From (1),
$\dot{s'} = k\dot{s}$ (5)
Putting (2) and (5) into (4):
$\frac{\dot{f}}{f^2} = -\frac{v}{s^2}\left(1+\frac{1}{k}\right)$ (6)
From (3), and using (1),
$\frac{1}{f} = \frac{1}{s}\left(1+\frac{1}{k}\right)$ (7)
Substituting (7) into (6),
$\dot{f} = \frac{df}{dt} = \frac{-vk}{k+1}$
the minus sign means that camera must reduce the focal distance at that rate.